Reaction between nitrogen and oxygen takes place as following:
`2N_(2(g))+O_(2)hArr2N_(2)O_((g))`
If a mixture of `0.482 "mole" N_(2)` and `0.933 "mole"` of `O_(2)` is placed in a reaction vessel of volume `10 litre` and allowed to form `N_(2)O` at a temperature for which `K_(c)=2.0xx10^(-37)litre mol^(-1)`. Determine the composition of equilibrium mixture.

Explanation

`[N_(2)]=(n)/(V)=(0.482)/(10)`
`=0.0482 mol litre^(-1)`
`[O_(2)]=(n)/(V)=(0.933)/(10)`
`=0.0933 mol litre^(-1)`
`{:(,,2N_(2),+,O_(2),hArr,2N_(2)O),(,"Initial conc.",0.0482,,0.0933,,0),(,"At equilibrium conc.",(0.0482-2x),,(0.0933-x),,2x):}`
`K_(c)=([N_(2)O]^(2))/([N_(2)]^(2)[O_(2)])=2xx10^(-37)`
`2xx10^(-37)=([N_(2)O]^(2))/((0.0482-2x)^(2)(0.0933-x))`
(`:' x` is very small)
`:. [N_(2)O]^(2)=(2xx10^(-37))(0.482)^(2)(0.0932)`
`=4.34xx10^(-41)`
`:. [N_(2)O]=6.6xx10^(-21) mol litre^(-1)`